1. Mass Relationships in Chemical Reactions Chapter 3

2. 화학양론 1. 원자질량 2.Avogadro 수와 몰질량 3. 분자질량 4. 질량분석기 5. 화합물의 조성백분율 6. 실험식 구하기 7. 화학반응과 반응식 8. 반응물과 생성물의 양 9. 한계반응물 10. 반응의 수득률

3. By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams

4. Natural lithium is: 7.42% 6 Li (6.015 amu) 92.58% 7 Li (7.016 amu) 7.42 × 6.015 + 92.58 × 7.016 100 = 6.941 amu Average atomic mass of lithium:

5. Average atomic mass (6.941)

6. The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C 1 mol = N A = 6.02214179 × 10 23 Avogadro’s number (N A ) The Mole(chemical quantity)

7. 겁 ( 劫 ) 과 찰라 ( 刹那 ) 찰나 [ 刹那 < 범어 k¡a¡a] ① ≪불교≫ (75 분의 1 초에 해당하는 ) 지극히 짧은 시간. 겁 ( 劫 ). ② 어떤 사물 현상이 이루어지는 바로 그때. 순간 ( 瞬間 ). 겁 [ 劫 ][< 범어 kalpa] ≪불교≫ (‘ 천지가 개벽한 때부터 다음 개벽할 때까지의 동안 ’ 이라는 뜻으로 ) ‘ 지극히 길고 오랜 시간 ’ 을 일컫는 말. 찰나. 힌두교에서는 1 겁을 43 억 2000 만 년 (1.36 × 10 17 초 ) 이라고 한다. =1.02 × 10 19

8. The Mole(chemical quantity) Avogadro’s number =6.022  10 23 만약 남북한 땅 위에 10 원짜리 동전을 쌓는다면 그 높이는 얼마나 될까 ? 10 직경 23 mm 두께 1.5 mm 녹여 쌓으면 1,700 km 사방으로 펼쳐 쌓으면 2,200 km 삼각으로 펼쳐 쌓으면 1,900 km

9. The Mole(chemical quantity) Avogadro’s number =6.022  10 23 만약 10 원 동전을 지구상 ( 땅과 바다 위 ) 에 쌓는다면 그 높이는 얼마나 될까 ? 녹여 쌓으면 730 m 사방으로 펼쳐 쌓으면 930 m 삼각으로 펼쳐 쌓으면 800 m 직경 23 mm 두께 1.5 mm

10. Molar mass is the mass of 1 mole of in grams atoms watermelons 1 mole 12 C atoms = 6.022 × 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole 12 C atoms = 12.00 g 12 C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

11. The Mole(chemical quantity) 1 몰 을 쌓자 ! 6.02 x 10 24 kg 지구 질량 5.97 × 10 24 kg 10 kg 수박

12. One Mole of: C S Cu Fe Hg

13. 1 amu = 1.66 × 10 -24 g or 1 g = 6.022 × 10 23 amu 1 12 C atom 12.00 amu × 12.00 g 6.022 × 10 23 12 C atoms = 1.66 × 10 -24 g 1 amu M = molar mass in g/mol N A = Avogadro’s number

14. Q. Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 × 10 23 atoms K × 6.022 × 10 23 atoms K 1 mol K = 0.551 g K 1 mol K 39.10 g K × 8.49 × 10 21 atoms K

15. Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S32.07 amu 2O + 2 × 16.00 amu SO 2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2

16. Q. Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C 3 H 8 O ? 5.82 x 10 24 atoms H 1 mol C 3 H 8 O = (3 × 12) + (8 × 1) + 16 = 60 g C 3 H 8 O 1 mol H = 6.022 × 10 23 atoms H 1 mol C 3 H 8 O molecules = 8 mol H atoms 72.5 g C 3 H 8 O 1 mol C 3 H 8 O 60 g C 3 H 8 O × 8 mol H atoms 1 mol C 3 H 8 O × 6.022 × 10 23 H atoms 1 mol H atoms × =

17. Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl NaCl

18. Q. Do You Understand Formula Mass? What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 × 40.08 2 P 2 × 30.97 8 O + 8 × 16.00 310.18 amu

19. KE = ½ × m × v 2 = qV v = (2 × qV/m) 1/2 F = q × v × B = m v 2 /R R =(mV/qB 2 ) 1/2 Light Heavy

20. Percent composition of an element in a compound = n × molar mass of element molar mass of compound × 100% n is the number of moles of the element in 1 mole of the compound C2H6OC2H6O %C = 2 × (12.01 g) 46.07 g × 100% = 52.14% %H = 6 × (1.008 g) 46.07 g × 100% = 13.13% %O = 1 × (16.00 g) 46.07 g × 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

21. Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. n K = 24.75 g K × = 0.6330 mol K 1 mol K 39.10 g K n Mn = 34.77 g Mn × = 0.6329 mol Mn 1 mol Mn 54.94 g Mn n O = 40.51 g O × = 2.532 mol O 1 mol O 16.00 g O

22. Percent Composition and Empirical Formulas K : ~ ~ 1.0 0.6330 0.6329 Mn : 0.6329 = 1.0 O : ~ ~ 4.0 2.532 0.6329 n K = 0.6330, n Mn = 0.6329, n O = 2.532 KMnO 4

23. g CO 2 mol CO 2 mol Cg C g H 2 O mol H 2 Omol Hg H g of O = g of sample – (g of C + g of H) Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O 6.0 g C = 0.5 mol C 1.5 g H = 1.5 mol H 4.0 g O = 0.25 mol O Empirical formula C 0.5 H 1.5 O 0.25 Divide by smallest subscript (0.25) Empirical formula C 2 H 6 O

24. 3 ways of representing the reaction of H 2 with O 2 to form H 2 O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproducts

25. How to “Read” Chemical Equations 2 Mg + O 2 2 MgO 2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O 2 makes 2 g MgO

26. Balancing Chemical Equations 1.Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water a C 2 H 6 + b O 2 c CO 2 + d H 2 O 2.Set the equation for the coefficients (a, b, c & d) to make the number of atoms of each element the same on both sides of the equation. C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d.

27. Balancing Chemical Equations 3.Solve the equations. Usually start by setting the coefficient of the most complex compound be 1. C: 2 a = c, O: 2 b = 2 c +d, H: 6 a = 2 d. 2 = c, 2 b = 2 c +d, 6 = 2 d. => d = 3 Let a = 1 2b = 2 × 2 + 3 = 7. b = 7/2 a =1 b = 7/2 c = 2 d = 3

28. Balancing Chemical Equations 4.Adjust the coefficient to the most simple whole numbers. a =1 b = 7/2 c = 2 d = 3 a =2 b = 7 c = 4 d = 6 × 2 2 C 2 H 6 + 7 O 2 4 CO 2 + 6 H 2 O

29. Balancing Chemical Equations 5.Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O ReactantsProducts 4 C 12 H 14 O 4 C 12 H 14 O 4 C (2 × 2) 4 C 12 H (2 × 6)12 H (6 × 2) 14 O (7 × 2)14 O (4 × 2 + 6)

30. 1.Write balanced chemical equation 2.Convert quantities of known substances into moles 3.Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4.Convert moles of sought quantity into desired units Amounts of Reactants and Products

31. Q. Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH × 4 mol H 2 O 2 mol CH 3 OH × 18.0 g H 2 O 1 mol H 2 O × = 235 g H 2 O

32. Limiting Reagents ( 한계반응물 ) 2NO + O 2 2NO 2 NO is the limiting reagent O 2 is the excess reagent NO: 7/2=3.5 O 2 : 7/1 =7 NO 2 : 3.5 × 2 =7 2NO + 1O 2 2NO 2

33. Q. Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. < Al is limiting reagent 124 g Al 2 27.0 g/mol Al = 2.296 601 g Fe 2 O 3 1 160.0 g/mol Fe 2 O 3 = 3.756

34. Use limiting reagent (Al) to calculate amount of product that can be formed. 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 2.296 × 102. g Al 2 O 3 = 234 g Al 2 O 3

35. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield × 100 Reaction Yield

36. Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3H 2 (g) + N 2 (g) 2NH 3 (g) NH 3 (aq) + HNO 3 (aq) NH 4 NO 3 (aq) 2Ca 5 (PO 4 ) 3 F (s) + 7H 2 SO 4 (aq) 3Ca(H 2 PO 4 ) 2 (aq) + 7CaSO 4 (aq) + 2HF (g) fluorapatite

37. 2H 2 + 1O 2 2H 2 O 2 분량 2.0g H 2 1 분량 32.0 g O 2 2 분량 18.0g H 2 O 2 mol H 2 1 mol O 2 2 mol H 2 O